Question

IF y$\sqrt{1-x^2}$+x$\sqrt{1-y^2}$=1 , Then prove that $\frac{dy}{dx}$=-$\sqrt{\frac{1-y^2}{1-x^2} }$

Answer 1

Answer:

√(1-y²)

- ------------

√(1-x²)

Step-by-step explanation:

given ---->

----------

y√(1-x²) + x√(1-y²)=1

To prove --->

---------------

dy √(1-y²)

--------- = --------------

dx √(1-x²)

proof --->

---------

y√(1-x²) + x √(1-y²)=1

let x=sinα , y =sinβ

=> α=sin-¹x , β=sin-¹y

sin β √(1-sin²α)+sinα √(1-sin²β)=1

we know that 1-sin²α=cos²α

and 1- sin²β=cos²β

using this

sinβ √cos²α +sinα √cos²β=1

sinβ cos α + sinα cosβ =1

we have a formula

sin (x+y)=sinx cosy+cosx siny

applying this

=> sin(β+α) = 1

=> β + α =1

=> sin-¹y + sin-1x=1

differentiating with respect to x

d d d

=> ------(sin-¹y) +-------(sin-¹X) =-------(1)

dx dx dx

1 d 1

=>--------------- ------(y) + ---------------=0

√(1-y²) dx √1-x²

1 dy 1

=> -------------- ------- = - -------------

√(1-y²) dx √1-x²

dy √1-y²

=> ------------ =- --------------

dx √1-x²

Hope it helps you

Thanks for giving me chance to answer your question

Answer 2

Answer:

Step-by-step explanation:

√(1-y²)

- ------------

√(1-x²)

Step-by-step explanation:

given ---->

----------

y√(1-x²) + x√(1-y²)=1

To prove --->

---------------

dy √(1-y²)

--------- = --------------

dx √(1-x²)

proof --->

---------

y√(1-x²) + x √(1-y²)=1

let x=sinα , y =sinβ

=> α=sin-¹x , β=sin-¹y

sin β √(1-sin²α)+sinα √(1-sin²β)=1

we know that 1-sin²α=cos²α

and 1- sin²β=cos²β

using this

sinβ √cos²α +sinα √cos²β=1

sinβ cos α + sinα cosβ =1

we have a formula

sin (x+y)=sinx cosy+cosx siny

applying this

=> sin(β+α) = 1

=> β + α =1

=> sin-¹y + sin-1x=1

differentiating with respect to x

d d d

=> ------(sin-¹y) +-------(sin-¹X) =-------(1)

dx dx dx

1 d 1

=>--------------- ------(y) + ---------------=0

√(1-y²) dx √1-x²

1 dy 1

=> -------------- ------- = - -------------

√(1-y²) dx √1-x²

dy √1-y²

=> ------------ =- --------------

dx √1-x²